Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 3}{x + 8} = \dfrac{-10x - 21}{x + 8}$
Solution: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 3}{x + 8} (x + 8) = \dfrac{-10x - 21}{x + 8} (x + 8)$ $ x^2 + 3 = -10x - 21$ Subtract $-10x - 21$ from both sides: $ x^2 + 3 - (-10x - 21) = -10x - 21 - (-10x - 21)$ $ x^2 + 3 + 10x + 21 = 0$ $ x^2 + 24 + 10x = 0$ Factor the expression: $ (x + 6)(x + 4) = 0$ Therefore $x = -6$ or $x = -4$ The original expression is defined at $x = -6$ and $x = -4$, so there are no extraneous solutions.